I am used to u-substitution integrals, where all $x$ cancel out, here it does not.
$\int x(x^2 + 1)^7 x \,dx$
Substituting, $u=x^2+1$.
Now, getting $dx$ with respect to $du$,
$\frac{du}{dx}=2x$
$dx=\frac{du}{2x}$
Putting this under the integral sign,
$\int xu^7x \frac{du}{2x}= \frac{1}{2}x\frac{u^8}{8}+c $ (one x now cancels out, one stays)
Now substituting back,
$= \frac{1}{2}x\frac{(x^2+1)^8}{8}+c = \frac{1}{16}x(x^2+1)^8+c$
I cannot understand which integral the OP wants to calculate. Does he want to calculate $\;\displaystyle\int x\left(x^2+1\right)^7\mathrm dx\;\;$ or $\;\displaystyle\int x^2\left(x^2+1\right)^7\mathrm dx\;\;?$
So I will calculate both integrals.
By letting $\;u=x^2+1\;,\;$ we get that $\;\mathrm du=2x\,\mathrm dx\;,\;$ hence ,
$\displaystyle\int\!x\left(x^2\!+\!1\right)^7\!\mathrm dx=\!\!\int\!\dfrac12\,u^7\mathrm du=\dfrac1{16}\,u^8\!+\!C=\dfrac1{16}\left(x^2\!+\!1\right)^8\!\!+C\,.$
$\displaystyle\int x^2\left(x^2+1\right)^7\mathrm dx=\displaystyle\int x^2\sum\limits_{n=0}^7\binom 7nx^{2n}\,\mathrm dx=$
$=\!\!\displaystyle\int\!x^2\!\left(1\!+\!7x^2\!+\!21x^4\!+\!35x^6\!+\!35x^8\!+\!21x^{10}\!+\!7x^{12}\!+\!x^{14}\right)\mathrm dx\!=$
$=\displaystyle\!\!\!\int\!\left(x^2\!+\!7x^4\!+\!21x^6\!+\!35x^8\!+\!35x^{10}\!+\!21x^{12}\!+\!7x^{14}\!+\!x^{16}\right)\mathrm dx\!=$
$=\!\dfrac13x^3\!\!+\!\dfrac75x^5\!\!+\!3x^7\!\!+\!\dfrac{35}9x^9\!\!+\!\dfrac{35}{11}x^{11}\!\!+\!\dfrac{21}{13}x^{13}\!\!+\!\dfrac7{15}x^{15}\!\!+\!\dfrac1{17}x^{17}\!\!+\!C$